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An aquarium 5 ft long, 2 ft wide, and 3 ft deep is full of water. Find (a) the hydrostatic pressure on the bottom of the aquarium, (b) the hydrostatic force on the bottom, and (c) the hydrostatic force on one end of the aquarium.

(a) $$

P=187.5 \mathrm{lb} / \mathrm{ft}^{2}

$$

(b) Force at the bottom of Aquarium $=1875 \mathrm{lb}$

(c) Hydro-static force on one end of the Aquarium is 562.5 $\mathrm{lb}$

Applications of Integration

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in this problem, we're given that there's an aquarium that is filled with order ended. Aquarium is five feet along, two feet wide and three feet, uh, deep. And where is to find the pressure at the bottom? We know that pressures equal to density times to gravitational acceleration. Times that and we can actually write this one as Linda de or Linda is the specific weight, which is just wrote 10 Jeep. And we know that Linda for water is 62.5 hell, part feet cube. So we need to do is to pluck the gnomes and soapy pressure will then be 62.5 supplied by that so we find pressure than to be £187.5 for feet. Skirt in part, Be worse. Find the force at the bottom of the tank. You know that forces you go to p times that where a is area. That pressure is acting more. In that case, a Buddha meeting area off this shaded region that would not be 1 87.5 times five feet. Tips to feed. We never find me before said to bottom of the aquarium to be one uh, £875 and important. See, whereas to find the force at one end. So you matter. We're looking at this aquarium from this side. So we would see Does that is two feet wide and 3/5 deep. Ah, let's assume not be having a tin stripped this guy unless you should read this City X. So the area Oh, or the force would then be p time, say or area is the main where your love interest That would be two times DX and you would need to done at those areas up starting from zero up to three, so area would be two times D X pressure again will be Lunda times deep. Ah, word that in this case, this unit is X. Let's say that we're measuring that ex starting from the surface off the water. So don't be alarmed. A Times X combining those two force would then be integral from 0 to 3. So serve some tea. Uh, thank the aquarium thio along up to the bottom of this aquarium. London Times X Times two D x. From this, we see that done force would be under X squared from 0 to 3 would be alumna times nine minus limit zero size nine. Linda, what we know about that is that gonna 62.5 pound for a few cute for water. So we find out the answer to be 5 62.5